（我被这道题恶心了大半个上午）

这道题是一个带mod的高斯消元，所以很明显不能用double类型的数组求解。

那么我们就需要在高斯消元的时候求出两个数的最小公倍数，然后分别乘上一个数以后相减。

接着依次判断无解，多解和只有一组解的情况，最后在求解的时候用乘法逆元（有一个小细节，就是ax=b时，a和b可能一正一负）或者exgcd。

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <map>
using namespace std;
const int N = 310, P = 7;
int power(int a, int b) {
    int ans = 1;
    while (b) {
        if (b & 1) ans = ans * a % P;
        a = a * a % P; b >>= 1;
    }
    return ans;
}
int n, m, a[N][N], ans[N];
map<string, int> mp;

void init() {
    mp["MON"] = 1;
    mp["TUE"] = 2;
    mp["WED"] = 3;
    mp["THU"] = 4;
    mp["FRI"] = 5;
    mp["SAT"] = 6;
    mp["SUN"] = 7;
}

int gcd(int a, int b) { return b != 0 ? gcd(b, a % b) : a;}

void Prepare() {
    int k, st, ed; string s;
    memset(a, 0, sizeof(a));
    for (int i = 1; i <= m; i++) {
        scanf("%d", &k);
        cin >> s; st = mp[s];
        cin >> s; ed = mp[s];
        a[i][n + 1] = (ed - st + 1 + P) % P;
        for (int j = 1; j <= k; j++) {
            scanf("%d", &st);
            a[i][st]++; a[i][st] %= P;
        }
    }
}

void Guass() {
    int now = 1; bool flag = 0;
    for (int i = 1; i <= n; i++) {
        int k = now;
        for (; k <= m; k++) if (a[k][i] != 0) break;
        if (k == m + 1) continue;
        if (k != now) for (int j = 1; j <= n + 1; j++) swap(a[now][j], a[k][j]);
        for (int j = 1; j <= m; j++) {
            if (j == now) continue;
            if (a[j][i] == 0) continue;
            int t = a[now][i] * a[j][i] / gcd(a[now][i], a[j][i]);
            int p1 = t / a[now][i], p2 = t / a[j][i];
            for (int k = 1; k <= n + 1; k++)
                a[j][k] = (a[j][k] * p2 - a[now][k] * p1) % P;
        }
        if (i == n) flag = 1;
        now++; if (now == m + 1) break;
    }
    for (int i = 1; i <= m; i++) {
        if (a[i][n + 1] == 0) continue;
        bool bk = 0;
        for (int j = 1; j <= n; j++)
            if (a[i][j] != 0) { bk = 1; break;}
        if (!bk) {puts("Inconsistent data."); return;}
    }
    if (!flag) {puts("Multiple solutions."); return;}
    for (int i = 1; i <= n; i++) {
        if (a[i][i] == 0) {puts("Multiple solutions."); return;}
        if (a[i][i] * a[i][n + 1] < 0) {
            if (a[i][i] < 0) a[i][i] = abs(a[i][i]), a[i][n + 1] = abs((a[i][n + 1] % P - P) % P);
            else a[i][n + 1] = (a[i][n + 1] % P + P) % P;
        }
        ans[i] = a[i][n + 1] * power(a[i][i], P - 2) % P;
        if (ans[i] < 3) ans[i] += P;
    }
    for (int i = 1; i <= n; i++) printf("%d ", ans[i]);
    puts("");
}

int main() {
    init();
    while (cin >> n >> m && n && m) {
        Prepare();
        Guass();
    }
    return 0;
}